Question

What is the relationship between time of flight $T$ and horizontal range $R$ ?
(where $\theta$ is angle of projection with the horizontal)

• A. $R=\frac{gT}{tan\, \theta}$
• B. $R=\frac{gT^{2}}{2\,tan\, \theta}$
• C. $R=\frac{gT^{2}}{tan\, \theta}$
• D. $R=\frac{gT}{2\,tan\, \theta}$

Answer 1

Answer: (B)

Horizontal range $R=\frac{u^{2}\,sin\,2\theta}{g}$
Time of flight $T=\frac{2u\, sin\,\theta}{g}$
where u is the velocity of the projection and $\theta$ is the angle of projection with the horizontal
$R=\frac{u^{2}\,sin\,2\theta}{g}$ or $R=\frac{2u^{2}\,sin\,\theta\,cos\,\theta}{g}$
or $ \frac{R}{2\,cos\,\theta} = \frac{u^{2}\,sin\,\theta}{g} \ldots\left(i\right)$
$T=\frac{2u\,sin\,\theta}{g}$ or $T_{2}=\frac{4u^{2}\,sin^{2}\,\theta}{g^{2}}$
$T^{2}=\frac{4}{g}\left(\frac{u^{2}\,sin\,\theta}{g}\right)sin\,\theta$
$=\frac{4}{g}\left(\frac{R}{2\,cos\,\theta}\right)sin\,\theta$ (Using (i))
or $T_{2} =\frac{2R}{g} tan\,\theta$
or $R=\frac{gT^{2}}{2\,tan\,\theta}$